Understanding exponent 00000000 and 11111111 in IEEE @doctopus: 1 11…111 • 2^-127 is closer to the previous (bigger) number than 0 11…111 • 2^-126 is, but note it was stated earlier that we decided to change the leading bit to 0 instead of decreasing the exponent
How to represent 11111111 as a byte in java - Stack Overflow When I say that 0b11111111 is a byte in java, it says " cannot convert int to byte," which is because, as i understand it, 11111111=256, and bytes in java are signed, and go from -128 to 127 But, if a byte is just 8 bits of data, isn't 11111111 8 bits?
Suppose you send data to the 11111111 11111111 11111111 11111111 IP . . . Suppose you send data to the 11111111 11111111 11111111 11111111 IP address on an IPv4 network To which device(s) are you transmitting? a All devices on the Internet b All devices on your local network c The one device that is configured with this IP address d No devices
why is 00000000 - 00000001 = 11111111 in C unsigned char data type? I observed that, when a unsigned char variable stores the value 0 (00000000 2) and it gets decremented by 1 (00000001 2), the variable value turns into 255 (11111111 2), which is the highest value that a unsigned char variable can hold My question is: why 00000000 2 - 00000001 2 turns into 11111111 2? (I want to see the arithmetic behind it)
c - addition of binary numbers - Stack Overflow Assuming two's complement (which is the most common representation and most likely what the question assumes), 11111111 is -1 and the sum will be 01000000, or 64 – Martin B Commented Sep 23, 2010 at 13:12
Answered: Consider the following bit stream- a. 00000000 b. 11111111 c . . . a 00000000 b 11111111 c 01010101 d 00110011 Draw the graph of the i NRZ-L ii NRZ-I iii Manchester iv differential Manchester scheme using each of the given data streams, assuming that the last signal level has been positive From the graphs, guess the bandwidth for this scheme using the average number of changes in the signal level